Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
#define OUT(x)                cout << #x << " = " << x << endl; 
#define rep(i, n)             for (int (i) = 0; (i) < (int)(n); (i)++)
#define rer(i, l, r)          for (int (i) = (int)(l); (i) <= (int)(r); (i)++)
#define reu(i, l, r)          for (int (i) = (int)(l); (i) < (int)(r); (i)++)
#define each(i, v)            for (auto i : v)
#define all(x)                (x).begin(), (x).end()
#define rall(x)               (x).rbegin(), (x).rend()
#define pb(x)                 push_back(x)
#define bp(x)                 __builtin_popcount(x)
#define mp(x, y)              make_pair((x), (y))
#define fi                    first
#define se                    second
#define setp(x)               setprecision(x)
#define mset(m, v)            memset(m, v, sizeof(m))
#define sz(x)                 (int)(x.size())
static const int INF        = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
static const int MOD        = 1000000007;
static const double PI      = 3.14159265358979;

#define int                   long long

typedef vector<double>        vd;
typedef vector<string>        vs;
typedef vector<bool>          vb;
typedef vector<int>           vi;
typedef pair<int, int>        pii;
typedef vector<pii>           vpii;

template<typename T> void pv(T a, T b) { for (T i = a; i != b; i ++) cout << *i << " "; cout << endl; }
template<typename T, typename U> inline void amin(T &x, U y) { if (y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if (x < y) x = y; }
int in() { int _x; scanf("%d", &_x); return _x; }
long long lin() {long long _x; scanf("%lld", &_x); return _x; };

vpii g[101010];
int dis[101010];
int sdis[101010];
int n;

void dijkstra(int s) {
        priority_queue<pii, vector<pii>, greater<pii>> q;
        fill(dis, dis + n, INF);
        dis[s] = 0;
        q.push(mp(s, 0));
        while (!q.empty()) {
                pii p = q.top(); q.pop();
                int v = p.fi;
                if (dis[v] < p.se) continue;
                each(e, g[v]) {
                        if (dis[e.fi] > dis[v] + e.se) {
                                dis[e.fi] = dis[v] + e.se;
                                q.push(mp(e.fi, dis[e.fi]));
                        }
                }
        }
}

signed main() { 
        int m;
        cin >> n >> m;
        int s, t;
        cin >> s >> t;
        s --;
        t --;
        rep(i, m) {
                int a, b;
                cin >> a >> b;
                a --;
                b --;
                g[a].pb(mp(b, 1));
                g[b].pb(mp(a, 1));
        }
        dijkstra(s);
        rep(i, n) sdis[i] = dis[i];
        dijkstra(t);
        int scnt[101010] = { 0 };
        int tcnt[101010] = { 0 };
        rep(i, n) {
                if (sdis[i] == INF) continue;
                scnt[sdis[i]] ++;
        }
        rep(i, n) {
                if (dis[i] == INF) continue;
                tcnt[dis[i]] ++;
        }
        scnt[0] = 1;
        tcnt[0] = 1;
        int k = sdis[t];
        k -= 2;
        int ans = 0;
        rer(l, 0, k) ans += scnt[l] * tcnt[k - l];
        cout << ans << endl;
        return 0;
}